原式=1/[n(n+1)]-1/[n(n+2)] =1/n-1/(n+1)-[1/2n-1/2(n+2)] =1/2n-1/(n+1)+1/2(n+2)
1/n(n+1)(n+2)=1/2[1/n(n+1)-1/(n+1)(n+2)] 这样才能多项 相消
1/n(n+1)(n+2)=0.5[1/n(n+1)-1/(n+1)(n+2)]
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